# Is the sum of white noise and shifted white noise white noise again?

Let $W[k]$ be a stationary white noise with variance = 1

Question: Is $X[k] = W[k] + c \cdot W[k-1]$ white noise?

$c$ is a real number.

• Spoiler: Lecture notes say it isn't but I can't see why... – Alon Mar 11 '18 at 18:23
• Hint: $W[k]+ cW[k-1]$ denotes a FIR filter. – Marcus Müller Mar 11 '18 at 19:05
• Abother option at proving this would really be just writing down the definition of the autocorrelation of a signal $Y[k]$, then inserting $X[k]=W[k]+cW[k-1]$ into that definition. You know the AKF of white noise! – Marcus Müller Mar 11 '18 at 19:06
• @MarcusMüller do you mind elaborating a little more on how I can derive the fact that this is not white noise by the fact that X[k] denotes a FIR filter? – Alon Mar 12 '18 at 10:12
• You ask yourself, "what's the result of applying that filter to a white signal, is it still white?", and notice that no, since that filter is for $c\ne0$ not an allpass, that's not the case. – Marcus Müller Mar 12 '18 at 10:23

Calculate the autocorrelation of the process.

\begin{align} R_{xx}[n] &=\mathbb{E}[(W[k] + c W[k-1])(W[k-n] + c W[k-1-n])] \\ &=\mathbb{E}[W[k]W[k-n]]+ \mathbb{E}[cW[k]W[k-1-n]]+\mathbb{E}[cW[k-1]W[k-n]]+\mathbb{E}[c^2W[k-1]W[k-1-n]] \\ &=\sigma^2\delta[n]+c\sigma^2\delta[n+1]+c\sigma^2\delta[n-1]+c^2\sigma^2\delta[n]\\ &=\sigma^2(1+c^2)\delta[n]+c\sigma^2\delta[n+1]+c\sigma^2\delta[n-1] \end{align}

The definition of white noise implies that $R_{xx}[n]=\sigma^2\delta[n]$, which is not the case here.

• Something is awry in this answer because the LHS is a function of $k$ while the RHS does not depend on $k$ at all after the second step. Furthermore, even if we take $n$ to be a typo for $k$, the alleged autocorrelation function is not an even function of its argument. – Dilip Sarwate Mar 12 '18 at 1:37
• No time to derive right now but the correlation of $X$ at lag k is zero. The only non-zero correlation is at lag one and it's equal to $c^2 \sigma^2$. $X$ is essentially an MA(1). – mark leeds Mar 12 '18 at 3:59
• @markleeds Note that the only requirement for autocorrelation and autocovariance to be equal is that the mean is $0$. By definition, $$C_{xx}(k)=\mathbb{E}[(X(n)-\mu_x)(X(n-k)-\mu_x)]$$ $$R_{xx}(k)=\mathbb{E}[X(n)X(n-k)]$$ If $\mu_x=0$, both are the same. – Tendero Mar 12 '18 at 15:51
• @Tendero, (continued) DSP frequenty (pun intended) deals with stationary signals where the means are zero, but assuming they are zero and altering the definition of correlation/covariance based on this assumption are two different matters. It certainly explains your recollection though. As more of a mathematician than an engineer, I find DSP folks are often a bit sloppy. Sometimes this is pragmatic, other times it introduces misunderstandings. – Cedron Dawg Mar 14 '18 at 16:45
• Hi Tendero, Cedron: When you write the covariance as an expectation, that E() has units and you don't want a "similarity" measure to be dependent on the units being used. So, correlation is just a scaling used to standardize the measure. Example:The t-stat ( another unitless measure ) in a simple linear regression can be written as a function of the correlation^2.. not covariance.^2. So, definitely not interchangeable in stats-land. Thanks to both of you for all the wisdom you provide to this list. – mark leeds Mar 14 '18 at 17:30